## n = 20

Âµ = 7

Ïƒ = 2

(7-2(2), 7+ 2(2)

(Âµ-2Ïƒ, Âµ + 2Ïƒ)

k = 2

1-1/k^2

1- 1/(2)^2

1-1/4

(4-1)/4 = 3/4 = 0.75

at least (20)(.75) = 15 shuttle flights lasted between 3 days and 11 days.

Kuai
answered

10 years ago

10 years ago

Âµ = 7

Ïƒ = 2

(7-2(2), 7+ 2(2)

(Âµ-2Ïƒ, Âµ + 2Ïƒ)

k = 2

1-1/k^2

1- 1/(2)^2

1-1/4

(4-1)/4 = 3/4 = 0.75

at least (20)(.75) = 15 shuttle flights lasted between 3 days and 11 days.

Clown Bot
answered

1 year ago

1 year ago

Now, if we want to know how many flights lasted between 3 and 11 days, we can safely say that all of the flights within the range of 5 to 9 days will also fall within the range of 3 to 11 days. So, at least 75% of the flights will last between 3 and 11 days.

But hey, don't worry about those other 25% of flights that might be outside that range. Maybe they had some unexpected sightseeing stops along the way or got caught in interstellar traffic. Who knows? Space travel can be unpredictable sometimes!

Step-by-Step Bot
answered

1 year ago

1 year ago

In this case, we can use Chebyshev's theorem to determine the minimum number of flights that lasted between 3 and 11 days.

First, let's calculate the number of standard deviations away from the mean that 3 days and 11 days are:

For 3 days:

(3 - 7) / 2 = -4 / 2 = -2

For 11 days:

(11 - 7) / 2 = 4 / 2 = 2

Now, let's calculate the fraction of data that falls within those bounds using Chebyshev's theorem:

For the lower bound (3 days):

At least 1 - 1/k^2 = 1 - 1/2^2 = 1 - 1/4 = 3/4 = 0.75

For the upper bound (11 days):

At least 1 - 1/k^2 = 1 - 1/2^2 = 1 - 1/4 = 3/4 = 0.75

Based on Chebyshev's theorem, at least 75% (3/4) of the flights lasted between 3 and 11 days.

To determine the minimum number of flights, we multiply the percentage by the total number of flights (20):

0.75 * 20 = 15

Therefore, at least 15 of the 20 space shuttle flights lasted between 3 and 11 days according to Chebyshev's theorem.

Explain Bot
answered

11 months ago

11 months ago

Chebyshev's theorem states that for any set of observations, the proportion of data values within k standard deviations of the mean is at least 1 - 1/k^2, where k > 1.

In this case, we know that the mean length of the first 20 space shuttle flights is 7 days, with a standard deviation of 2 days. To find the range within 3 and 11 days, we need to calculate the number of standard deviations each value is from the mean.

First, let's calculate the number of standard deviations for each value:

- For 3 days: (3 - 7) / 2 = -2

- For 11 days: (11 - 7) / 2 = 2

Since we want to find the range between -2 and 2 standard deviations from the mean, where k > 1, we can use k = 2.

According to Chebyshev's theorem, at least 1 - 1/2^2 = 1 - 1/4 = 3/4 or 75% of the data values are within 2 standard deviations from the mean.

Using this information, we can conclude that at least 75% of the first 20 space shuttle flights lasted between 3 and 11 days.